插你个值
scturtle
posted @ 2010年4月05日 03:07
in 各种实验
, 1591 阅读
最近浑浑噩噩的 觉得还是打游戏比较有前途吧
lagrange 插值
function l
clc,clear,close
a=[-3,0,3,6];
b=[-1,2,-2,10];
for i=1:100
x(i)=i;
y(i)=lagrange(a,b,x(i));
end
plot(x,y)
function y=lagrange(a,b,x)
n=length(a);
y=0;
for i=1:n
y=y+b(i)*li(x,a,i);
end
function f=li(x,a,i)
n=length(a);
f=1;
for j=1:n
if j~=i
f=f*(x-a(j));
f=f/(a(i)-a(j));
end
end
newton插值
function newton
clc,clear,close
a=[-3,0,3,6];
b=[-1,2,-2,10];
for i=1:100
x(i)=i;
y(i)=f(a,b,x(i));
end
plot(x,y)
function y=f(a,b,x)
n=length(a);
y=b(1);
for i=2:n
t=1;
for j=1:i-1
t=t*(x-a(j));
end
y=y+t*juncha(a,b,1,i);
end
function f=juncha(a,b,m,n)
if m==n
f=b(m);
else
f=(juncha(a,b,m,n-1)-juncha(a,b,m+1,n))/(a(m)-a(n));
end
Aitken:
function a
clc,clear,close
a=[-3,0,3,6];
b=[-1,2,-2,10];
for i=1:100
x(i)=i;
y(i)=aitken(a,b,x(i));
end
plot(x,y)
function y=aitken(a,b,x)
n=length(a);
if n==2
y=b(1)*(x-a(2))-b(2)*(x-a(1));
y=y/(a(1)-a(2));
else
c=a;d=b;
c(n)=[];d(n)=[];
t1=aitken(c,d,x);
c=a;d=b;
c(n-1)=[];d(n-1)=[];
t2=aitken(c,d,x);
y=t1*(x-a(n))-t2*(x-a(n-1));
y=y/(a(n-1)-a(n));
end
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